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- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.Insurance Company A claims that its customers pay less for car insurance, on average than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 11 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $14. For 14 randomly selected customers of Company B, you find that they pay a mean of $158 per month with a standard deviation of $12. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.For a study conducted by the research department of a pharmaceutical company, 235 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.80 with a standard deviation of $4.30 . A random sample of 250 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.20 with a standard deviation of $4.00 . As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 90% confidence interval for the difference −μ1μ2 between the mean spending on prescription allergy relief medication ( μ1 ) and the mean…
- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.In a simple random sample of 144 households in a county in Virginia, the average number of children in these households was 3.62 children. The standard deviation from this sample was 2.40 children. A 90% confidence interval for the mean number of children in all households in this county is?a. 14.0, 16.3, 18.7, 19.9, 21.3 b. 16.1, 13.4, 13.6, 14.9, 14.1, 10.5 Construct the 95% confidence interval for the difference in u of the two given samples.
- According to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.A simple random sample of 5 observations from a normal distribution was taken, and the following values were obtained: {15, 19, 19, 20, 22}. A 95% confidence interval for the mean is O A. (16.569, 21.431) O B. (15.835, 22.165) O C.(16.069, 21.931) O D. (17.124, 20.876) O E. (16.765, 21.235)A researcher was interested in comparing the mean grade-point-average (GPA) of students at the University of Toronto Scarborough and Mississauga campuses. Independent random samples of 8 students from the Scarborough campus and 13 students from the Mississauga campus yielded the following GPAs. You can assume the data in each group follow a normal distribution. Do the data provide sufficient evidence to conclude that the mean GPA of students at the Scarborough campus different from the mean GPA of students at the Mississauga campus? A. What statistical test is most appropriate to answer this question? B. Identify the relevant assumptions and conditions. State how these have or have not been met. C. Write an appropriate null and alternative hypothesis for this question.
- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.we look at a random sample of 1000 united flights in the month of December comparing the actual arrival time to the scheduled arrive time. computer output of the descriptive statistics for the difference in actual and expected arrival time of these 1000 flights are shown below : N=1000 Mean =4.06 St Dev. = 45.4 SE Mean =1.44 Min= -35 Q1 = -14 Median = -5 Q3 =9 Max =871 what is the sample mean difference in actual and expected arrival times? what is the standard deviation of the difference ? x bar = and s= ?2. A sample of six college basketball player had an average weight of 217 pounds with a sample standard deviation of 10 pounds. Find the 95% confidence interval of the true mean weight of all basketball player.